Question

If $\left( {1 + 3 + 5 + .... + p} \right) + \left( {1 + 3 + 5 + .... + q} \right) = \left( {1 + 3 + 5 + .... + r} \right)$, where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of $p + q + r$, (where $p > 6$ ) is:
A. 12
B. 21
C. 27
D. 24

Answer 1

To solve this question first we find that this expression is in arithmetic progression then find the sum of all the three expressions of then add them and try to make in the form of Pythagoras triplet and then we assign the smallest Pythagoras triplet and then find the value of all the variables that we need and add them to get the final answer.

Complete step by step answer:
We have given that $\left( {1 + 3 + 5 + .... + p} \right) + \left( {1 + 3 + 5 + .... + q} \right) = \left( {1 + 3 + 5 + .... + r} \right)$.
We have to find the then minimum value of $p + q + r$.
$\left( {1 + 3 + 5 + .... + p} \right) + \left( {1 + 3 + 5 + .... + q} \right) = \left( {1 + 3 + 5 + .... + r} \right)$
All the three expression are in arithmetic progression so we use the formula of sum of n term of arithmetic progression.The sum of first expression is:
$1 + 3 + 5 + .... + p = \dfrac{{{{\left( {p + 1} \right)}^2}}}{4}$

The sum of second expression is:
$1 + 3 + 5 + .... + q = \dfrac{{{{\left( {q + 1} \right)}^2}}}{4}$
The sum of third expression is:
$1 + 3 + 5 + .... + r = \dfrac{{{{\left( {r + 1} \right)}^2}}}{4}$
Now putting all the values in the $\left( {1 + 3 + 5 + .... + p} \right) + \left( {1 + 3 + 5 + .... + q} \right) = \left( {1 + 3 + 5 + .... + r} \right)$ in form of sum.
$\dfrac{{{{\left( {p + 1} \right)}^2}}}{4} + \dfrac{{{{\left( {1 + q} \right)}^2}}}{4} = \dfrac{{{{\left( {1 + r} \right)}^2}}}{4}$

Now canceling the factor of 4 from both sides.
${\left( {p + 1} \right)^2} + {\left( {1 + q} \right)^2} = {\left( {1 + r} \right)^2}$
The obtained equation is a Pythagoras triplet
$p + 1,q + 1,r + 1 \in \;Pythagoras{\text{ }}triplet$
$p > 6$ so we take $p = 7$ and the smallest possible value of $q,r$.
$p = 7$, $q = 5$ and $r = 9$ these are the smallest values of all variables.
The smallest possible of $p + q + r$ is $7 + 5 + 9 = 21$

Hence, the correct answer is option B.

Note: To solve these types of questions students have knowledge of types of progression and must know all the formulas related to those expressions. Students often make mistakes in deciding the value of the variable and then don’t apply the condition on one variable and then directly equate all those to 3,4,5 in order to find the smallest sum.